Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{7k^3 + 21k^2 + 14k}{-3k^2 + 3}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {7k(k^2 + 3k + 2)} {-3(k^2 - 1)} $ $ a = -\dfrac{7k}{3} \cdot \dfrac{k^2 + 3k + 2}{k^2 - 1} $ Next factor the numerator and denominator. $ a = - \dfrac{7k}{3} \cdot \dfrac{(k + 1)(k + 2)}{(k + 1)(k - 1)}$ Assuming $k \neq -1$ , we can cancel the $k + 1$ $ a = - \dfrac{7k}{3} \cdot \dfrac{k + 2}{k - 1}$ Therefore: $ a = \dfrac{ -7k(k + 2)}{ 3(k - 1)}$, $k \neq -1$